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Counting permutations uses combinatorics and factorials

Example: For \( n \) items, the number of permutations is equal to \( n! \) (factorial of \( n \))

How to count distinct permutations?

Having a repeated item involves a division of the number of permutations. Count the number of permutations of these repeated items.

Example: DCODE letters have \( 5! = 120 \) permutations but contain the letter D twice (these \( 2 \) letters D have \( 2! \) permutations), so divide the total number of permutations \( 5! \) by \( 2! \): \( 5!/2!=60 \) distinct permutations.

How to remove the limit when computing permutations?

Permutations makes exponential values witch needs huge computing servers with huge memory cells, so the generation must be paid.

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